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The Mixture measurements and the Gibbs-Duhem constraint article describes the inspected acetone–chloroform preparation, apparatus, paired readouts, controls and sampling limits.^[ref]

That existing apparatus account supplies the experimental controls and limitations. Its refractive index calibration yields mass fractions, which must be converted to mole fractions. For a measured component-A mass fraction $w_A$ and component molar masses $M_A,M_B$,Apply this conversion separately to the flask sample and to completely condensed vapor to obtain $x_A$ and $y_A$. Comparing $y_{i}P$ with $x_i{p}_i^{\ast }(T)$ tests Raoult's law at the measured temperature; using the changing temperature at each composition is essential. The chosen A/B values above are unrelated to the protocol's substances.

The more general diagnostic, retaining an ideal gas vapor and negligible pure-liquid pressure correction, isThis follows by replacing $\ln x_i$ with $\ln (x_i{\gamma }_i)$ in the chemical potential subtraction used for Raoult's law. A reproducible discrepancy from ${\gamma }_i=1$ can expose failure of the ideal solution approximation, but interpreting it requires checking measurement uncertainty, representative paired sampling and the ideal gas approximation. Apparent agreement within uncertainty supports an approximation over the tested conditions; it cannot establish exact ideality at every composition.

Positive or negative deviations of component partial pressures require activity coefficients different from one under these assumptions. An azeotrope is also absent from the chosen unequal-vapor pressure model: the explicit inequality in ideal binary vapor composition gives $y_A>x_A$ throughout $0<x_A<1$. An interior equality would require $p_A^{\ast }=p_B^{\ast }$; in that degenerate case every composition has $y_A=x_A$, rather than an isolated enrichment reversal. Neither an azeotrope nor liquid phase separation should be inferred from the ideal curves.

For liquid–gas phase coexistence, make two distinct approximations: the gas obeys the ideal gas law, $v_{m,g}=RT/p$, and its molar volume greatly exceeds the liquid molar volume, $v_{m,g}\gg v_{m,l}$. Here $R$ is the molar gas constant and $p$ is the saturation pressure of the pure substance. Substitution into the Clapeyron equation giveswhere $p_{\ast }>0$ is any fixed reference pressure, making the logarithm dimensionless. Constant latent heat is a further approximation, needed for the elementary Integrated Clausius-Clapeyron relation below.^[ref]

The neglected liquid molar volume has a transparent error estimate. If the gas is ideal but $ϵ=v_{m,l}/v_{m,g}$ is retained, thenThus ignoring the liquid molar volume underestimates this slope by a relative amount $ϵ$. This estimate does not bound errors from treating a real gas as ideal or from freezing $L_m(T)$. Near a thermodynamic critical point, the gas and liquid molar volumes approach each other and these dilute-vapor assumptions fail.

If the molar latent heat $L_m$ is approximately constant on the chosen temperature interval, the ideal-vapor Clausius-Clapeyron approximation integrates toFor $L_m>0$, higher absolute temperature gives higher saturation pressure. A plot of $\ln (p/p_1)$ against inverse temperature $1/T$ is a straight line of slope $-L_m/R$. One reference pair $(T_1,p_1)$ fixes its intercept. On a per-unit-mass basis replace $L_m/R$ by $L/R_s$, with $R_s=R/M$ and $M$ the molar mass; using $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ with the molar $R$ would give incompatible units.

If only the constant-latent heat approximation is relaxed, the same ideal-vapor model instead givesConsequently an apparent straight-line fit across a broad temperature range need not represent a single material constant. The simple exponential also approaches a finite value as $T\to \infty$; this is a property of the stipulated formula, not a prediction that liquid–gas phase coexistence extends beyond a thermodynamic critical point.

On 10 July 1908, Heike Kamerlingh Onnes's Leiden laboratory circulated purified helium through liquid air and evaporatively cooled liquid-hydrogen precooling, a regenerative heat exchanger, and an expansion valve. Returning helium cooled incoming helium; liquid collected in a transparent vacuum vessel. Reflected light revealed its liquid meniscus, crossed by thermocouple wires. Subsequent mass density and spectroscopy checks tested contamination.^[ref]

A constant-volume helium gas thermometer below the valve supplied the temperature indication. An oxygen boiling reference check disagreed with its expected reading; the real gas correction to thermodynamic temperature remained provisional. A temperature plateau suggested immersion before the liquid meniscus was recognized.^[ref]

The measurement principles separate two questions. A visible liquid meniscus establishes an interface; a gas thermometer estimates temperature through an equation of state. A stationary temperature trace can also result from a steady state without phase coexistence, so it cannot replace the interface observation. Conversely, seeing a liquid does not calibrate a thermometer. Thermometer calibration, the gas thermometer dead-space correction, departures from the ideal gas law and incomplete thermal equilibrium all matter when assigning an absolute temperature. A check far from the intended measurement range tests the instrument but does not by itself bound every low-temperature error.

A laboratory condenser removes heat from vapor so that it becomes liquid. In distillation, the downstream laboratory condenser makes collectable distillate. Complete condensation, without loss or reaction, preserves the entering vapor's overall component mole fractions; cooling alone does not give another enrichment step. Partial condensation can instead leave two streams of different composition, requiring both vapor-liquid equilibrium and a distillation material balance.

The lambda point is the temperature of the transition between ordinary liquid helium-4 and its superfluid phase, named after the sharp lambda-shaped heat capacity feature. At saturated vapor pressure it lies near $2.17\mathrm{K}$. Its value varies with pressure.

The Landau criterion asks whether creating an excitation can lower a moving superfluid's energy. For a weakly coupled, heavy obstacle moving through a uniform liquid initially at zero temperature, let $ϵ(p)>0$ be the excitation energy in the fluid rest frame, with momentum magnitude $p$. Transferring momentum $\mathbf{p}$ from an obstacle moving at velocity $\mathbf{v}$ supplies, to leading order in its negligible recoil, energy $\mathbf{v}\cdot \mathbf{p}$. Emission requiresFor an isotropic dispersion relation, the easiest direction is parallel to $\mathbf{v}$, giving the necessary kinematic conditionAn infimum includes cases in which the limiting slope occurs only as $p\to 0$. A finite-mass obstacle of mass $M$ instead loses $\mathbf{v}\cdot \mathbf{p}-p^2/(2M)$, so the corresponding emission inequality includes the extra positive recoil term $p^2/(2M)$ on the left. This is an energy and momentum balance, not a calculation of an emission rate.

The Landau criterion excludes spontaneous emission into the specified spectrum below its threshold. Above it, kinematic permission alone does not guarantee strong coupling or rapid decay. It is not a universal observed flow threshold: walls, channel geometry and quantum vortex nucleation can enable other decay routes, and finite temperature permits scattering from existing excitations. Using only bulk phonons and rotons leaves those processes outside the model.

Landau's 1941 energy and momentum argument used a roton approximation different from the finite-momentum minimum illustrated below.^[ref]

Landau received his Nobel Prize in Physics in 1962 while hospitalized in Moscow after a car accident.^[ref]

For a deliberately simple Bose-Einstein condensation comparison, stipulate an ideal free-particle excitation law $ϵ=p^2/(2m)$. Then $ϵ/p=p/(2m)$ has infimum zero. Macroscopic one-particle occupation therefore supplies no positive emission threshold by this argument alone. By contrast a linear phonon law $ϵ=cp$ yields $v_L=c$. The dispersion relation and the collective response contain information that the occupation criterion alone does not supply; Bose-Einstein condensation and thermodynamic superfluidity must be tested separately.

To check the picture algebraically, on the stipulated roton branch,Thus tangency occurs at $q_L=\sqrt{4.3}$, slightly above the roton energy minimum. At an energy minimum $e^{\prime }=0$, but a tangent through the origin needs $e^{\prime }=e/q>0$. Minimizing energy and minimizing the energy-to-momentum ratio are different operations. This geometric construction makes the criterion useful even when the full dispersion relation is supplied numerically rather than by a simple formula.

Latent heat is the heat absorbed or released in a first-order phase transition through its entropy jump. For an equilibrium conversion from thermodynamic phase $\alpha$ to thermodynamic phase $\beta$ at fixed pressure, evaluate both thermodynamic phases at the same temperature and pressure on their phase coexistence curve:Use one normalization throughout: $L$, $S$ and $H$ may all be per mole, giving $L$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or all per unit mass, giving $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Here $S$ is entropy and $H$ is enthalpy on that basis. Equal Gibbs free energies of the coexisting thermodynamic phases give $\Delta H=T_{\mathrm{tr}}\Delta S$. Identifying this enthalpy difference with supplied heat assumes a closed system and an equilibrium conversion with only pressure-volume work. ^[ref]

For a pure substance with positive latent heat per unit mass $L$, let $x$ be the phase fraction of the higher-enthalpy thermodynamic phase, measured by mass. While both thermodynamic phases coexist at the fixed pressure,Adding heat changes $x$ rather than the temperature; this creates the horizontal portion of a heating curve. Once $x=1$, additional heat raises the temperature according to specific heat capacity at constant pressure. Reversing the conversion releases heat. The plateau represents changing enthalpy, rather than a finite single-thermodynamic phase heat capacity satisfying $Q=C_p\Delta T$. ^[ref]

For a constructed fixed-pressure model, take $m=1\mathrm{kg}$, initial temperature $280\mathrm{K}$, transition temperature $T_{\mathrm{tr}}=300\mathrm{K}$, solid specific heat capacity at constant pressure $c_s=2\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, latent heat $L=100\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, and liquid specific heat capacity at constant pressure $c_l=3\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. These are stipulated model parameters, not empirical properties of a named material. Heating the solid to $300\mathrm{K}$ takes $40\mathrm{kJ}$ of sensible heat; converting it fully to liquid takes another $100\mathrm{kJ}$ of heat; heating the liquid to $320\mathrm{K}$ takes $60\mathrm{kJ}$. The total is $200\mathrm{kJ}$. After a total heat input of $90\mathrm{kJ}$, $x=(90-40)/100=1/2$ and the temperature is still $300\mathrm{K}$. The corresponding per-unit-mass entropy jump is $L/T_{\mathrm{tr}}=\frac{1}{3}\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$.

A continuous phase transition has no latent heat. Its heat capacity can have a jump or a singularity without a finite interval in which added heat converts coexisting thermodynamic phases at a constant temperature. A heating plateau therefore requires the stated phase coexistence conditions; it is not a feature of every phase transition. ^[ref]

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

Consider an original model of a cold section enclosing a throttle and collecting region, with the precooling heat exchanger outside its boundary. The inlet has mass flow rate $\dot{m}>0$ and specific enthalpy $h_{\mathrm{in}}$. The outlets are equilibrium liquid and vapor at one fixed pressure, with specific enthalpies $h_l$ and $h_g$. Let $y$ be the outlet liquid mass fraction, and let $q=\dot{Q}/\dot{m}\ge 0$ be incoming unwanted heat per unit incoming mass. Assume steady flow, no shaft work, and negligible bulk kinetic energy and gravitational potential energy changes. The steady-flow energy balance givesHere $L$ is the latent heat per unit mass at the chosen outlet pressure. The mass flow rates of the outlets are $y\dot{m}$ and $(1-y)\dot{m}$. The formula describes a flowing cold section; a filling vessel with appreciable energy or mass accumulation requires storage terms.

For a deliberately dimensionless example, choosePer unit incoming mass, one quarter leaves as liquid and three quarters as vapor. With zero incoming heat, the same inlet would give $y=0.30$. Increasing $q/L$ to $0.20$ reduces $y$ to $0.10$, and $q/L=0.30$ reaches the no-liquid boundary. These are chosen model values, not helium property data or a historical yield. In normalized enthalpy units with $h_l=0$ and $h_g=1$, conservation checks directly: $0.70+0.05=0.25\times 0+0.75\times 1$.

The admissible interval is $h_l\le h_{\mathrm{in}}+q\le h_g$, or $0\le y\le 1$. Outside it, the assumed equilibrium two-phase outlet fails: an enthalpy above $h_g$ requires a hotter single-vapor outlet under these assumptions, rather than a negative amount of liquid. Below $h_l$, a colder liquid or a different thermodynamic phase must be considered. The balance does not predict its own outlet pressure, temperature or helium property functions.

At fixed outlet pressure, hence fixed $L$, the derivatives are $\partial y/\partial h_{\mathrm{in}}=-1/L$ and $\partial y/\partial q=-1/L$. Lowering inlet specific enthalpy by regeneration and reducing incoming heat therefore buy the same increase in liquid mass fraction per unit energy per unit mass in this model. Adding any constant to all three specific enthalpies leaves $y$ unchanged. This independence of the arbitrary enthalpy reference makes the balance useful even when property tables choose different zero points.

Conversion of a gas to a liquid.

Apply the steady-flow energy balance to an entire liquefaction apparatus, allowing a warm gas outlet and a separate cold liquid outlet. With incoming heat $q$ per unit feed mass, no shaft work, and negligible changes in bulk kinetic energy and gravitational potential energy, conservation of mass and conservation of energy give $h_{\mathrm{in}}+q=y{h}_l+(1-y)h_g$. Solving this equation gives the displayed liquid mass fraction.

The two outlet streams need not share a temperature. For example, a counterflow heat exchanger can warm returning gas before it crosses the overall control volume boundary, while collected liquid leaves at the cold end. When the outlet temperatures differ, $h_g-h_l$ includes sensible heating as well as phase change; it is not simply a latent heat. A colder two-phase separator nested inside the apparatus has its own boundary states and a different local enthalpy balance.

A liquid that produces particle scintillation when deposited energy excites its components. Photomultipliers can convert the emitted light into detector pulses.

The liquid phase of water. Its density is much greater than that of dilute water vapor, allowing the liquid's specific volume to be neglected in the usual integrated Clausius-Clapeyron relation.

An instrument for measuring pressure relative to another pressure. A liquid-column manometer uses hydrostatic pressure; a digital manometer uses a pressure transducer.

The appeal of a liquid-column manometer is that a pressure measurement becomes a visible length measurement: mass density and gravitational acceleration supply the conversion. It can compare two pressure tappings in a flowing fluid, even though the measuring liquid itself must settle into hydrostatic equilibrium. The reference connection determines which pressure difference is being measured.

The ideal measurement calibration is a straight line through the equal-pressure zero. For the stipulated water mass density and gravitational acceleration above, its slope is $9.81\mathrm{Pa}$ per vertical millimetre. A lower liquid mass density gives a larger length change for the same pressure, improving the height-reading contribution to measurement resolution but requiring more vertical space for a given pressure range. The pressure scale depends on the actual mass density, which depends on temperature; $1000\mathrm{kg}{\mathrm{m}}^{-3}$ here is a chosen model value, not an exact property of all water samples.

Suppose each liquid meniscus height is rounded to the nearest scale division $q=1\mathrm{mm}$. Each height measurement error lies within $q/2$, so the error in their difference lies within $q$. The resulting pressure error is bounded by $\rho gq=9.81\mathrm{Pa}$. To quantify a standard uncertainty, make the additional model assumption that the two rounding errors are independent random variables, each having a continuous uniform distribution on $[-q/2,q/2]$. Each then has variance $q^2/12$; adding the two variances givesHere $u$ denotes standard uncertainty. The $9.81\mathrm{Pa}$ worst-case bound and $4.0049\mathrm{Pa}$ standard uncertainty answer different questions. At $588.6\mathrm{Pa}$ they are respectively $1.67\%$ and $0.680\%$ of the reading. Shared reading bias or a different reading method can violate the assumed independence.

If the mass density, gravitational acceleration and height difference also have small independent standard uncertainties, a first-order estimate follows from the partial derivatives of $\Delta p=\rho g\Delta h$:This displays why a finer scale does not remove uncertainty in mass density or gravitational acceleration. A pressure transducer compared against the column needs its own differential-pressure calibration to establish an electrical response.

A curved free interface of a liquid near a solid wall. Its shape reflects surface tension, contact angle and hydrostatic pressure.

A liquid meniscus is the curved boundary where a liquid meets a vessel wall. In the transparent water examples above, the bottom of this curve is aligned with the calibration mark at eye level. Reading from above or below changes the apparent alignment without changing the contained volume.

For a small heel of a floating body in a uniform fluid, the metacentre is the intersection of the new vertical buoyancy line with the original vertical through the centre of buoyancy. With the centre of mass a signed distance $BG$ above the original centre of buoyancy, the metacentric height is $GM=I_{\mathrm{wp}}/V_d-BG$. Here $I_{\mathrm{wp}}$ is the waterplane second moment of area about the heel axis and $V_d$ the displaced volume. Positive $GM$ gives a restoring small-angle torque and negative $GM$ gives instability. The formula describes local static stability, not all large-angle behaviour or effects of free liquid surfaces inside the body.

A mixture contains two or more chemical substances combined in proportions that need not be fixed. A chemical compound instead has a definite composition: water has two hydrogen chemical atoms per oxygen chemical atom, whereas a sucrose and water mixture can contain different sugar proportions. A homogeneous mixture has uniform composition on the scale of representative samples; a heterogeneous mixture has distinguishable regions of different composition. A chemical solution is homogeneous, whereas sand dispersed in water is a particle suspension.^[ref]

The distinction concerns the sample, not merely its ingredient list. A vessel with dissolved sodium chloride and excess undissolved sodium chloride contains a chemical solution and an ionic crystal thermodynamic phase. The liquid portion can be homogeneous while the whole vessel is heterogeneous. Similarly, molecular oxygen dissolved in water is part of a chemical solution; visible gas bubbles introduce a separate gas thermodynamic phase.